EE 313 Linear Signals & Systems (Fall 2018) Solution Set for Homework #8 on Continuous-Time Signals & Systems

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1 EE 33 Linear Signals & Sysems (Fall 08) Soluion Se for Homework #8 on Coninuous-Time Signals & Sysems By: Mr. Houshang Salimian & Prof. Brian L. Evans Here are several useful properies of he Dirac dela funcional (generalized funcion): a) Uni area: δ x dx = b) Sifing propery: f x δ x x dx = c) Even symmery: δ x = δ x f x, x [a, b] 0, oherwise d) Relaionship o he uni sep funcion. u x = δ x. " Here are several commens abou bounded-inpu bounded-oupu (BIBO) sabiliy: e) BIBO Sabiliy: If inpu x() is bounded in ampliude, i.e. x B for a finie value B, hen oupu y() is always bounded in ampliude, i.e. y B for a finie value B. This definiion does no require he sysem o be LTI. f) BIBO sabiliy for LTI sysems: For a coninuous-ime LTI sysem wih an impulse response h(), BIBO sabiliy reduces o h d <. A derivaion is given in problem 3 below. g) BIBO sabiliy for FIR filers: From f), i immediaely follows ha FIR filers are always BIBO sable (if h() < for all ). This is also refleced in he fac ha all he poles of an FIR filer are a z=0 (inside he uni circle), which implies sabiliy. Please see Handou I on Bounded-Inpu Bounded-Oupu Sabiliy a hp://users.ece.uexas.edu/~bevans/courses/signals/handous/appendix%0h%0bibo%0sabiliy.pdf Convoluion: Le c() = x()*y() => c = Problem : x τ y τ dτ = x τ y dτ a) In his quesion we can use he following propery: δ ( )* f( ) = f( ) 0 0 ( 0) ( ) e u( 0) cos(00π 000) ( 0) ( ) δ( 0)*[ δ( + 0) + e u( ) + cos(00 π)] = δ ( 0) e u( 0) + cos(00 π( 0)) = δ + + b) The Dirac dela funcional is defined in erms of inegraion: (a) i has uni area a he origin and (b) has a sifing propery. The Dirac dela funcional is waiing around o be inegraed. Please avoid simplifying expressions involving he Dirac dela ha are no being inegraed. [ ] ( ) cos(00 π) δ( ) + δ( 0.00) d = cos(0) δ( ) d+ cos 0. π δ( 0.00) d = + cos(0. π ) =.809 c) The Dirac dela funcional is defined in erms of inegraion: (a) i has uni area a he origin and (b) has a sifing propery. The Dirac dela funcional is waiing around o be inegraed. Please avoid simplifying expressions involving he Dirac dela ha are no being inegraed

2 Fall 08 EE 33 Homework 8 soluion The Universiy of Texas a Ausin d e ( ) u( ) d = d e ( ) d ( u( ) ) + u( ) d d ( e ( ) ) = e ( ) δ( ) e ( ) u( ) d) ( ) ( ) ( ) d ( ) cos 00πτ δ τ + δ τ 0.00 τ = cos 0 δ ( τ ) + cos(0. π ) δ ( τ 0.00) dτ ( ) ( ) = cos 0 δτ ( ) dτ+ cos 0. π δτ ( 0.00) dτ= u( ) + cos(0. π) u( 0.00) = u( ) u( 0.00) Problem : This is averaging filer (unnormalized). Is oupu is he average of he previous wo seconds of inpu, he curren inpu value, and he fuure wo seconds of inpu. If a gain of ¼ had been applied, hen we d have a normalized averaging filer (normalized so ha he area of he absolue value of he impulse response is one). + y () = x( τ) dτ a) + + h ( ) = δτ ( ) dτ= δτ ( ) dτ δτ ( ) dτ= δτ ( ʹ+ ) dτʹ δτ ( ʹʹ ) dτʹʹ= u ( + ) u ( ) Alernae Soluion: 0, + < h () = δτ ( ) dτ=, < 0, h () = u ( + ) u ( ) b) If x B for all, hen y = x τ dτ x τ dτ Bdτ = 4B So, a bounded inpu generaes a bounded oupu and hence he sysem is bounded-inpu bounded-oupu (BIBO) sable. A coninuous-ime LTI sysem is sable if and only if: Here, h(τ) dτ = dτ h(τ) dτ < = 4 and he sysem is BIBO sable. c) This sysem is no causal, because curren oupu is dependan o fuure value of inpu. For d) insance a =: y = in fuure, i.e. = o 3. x(τ)dτ which shows ha oupu a = is relaed o inpu values Noe: A coninuous-ime, LTI sysem is causal if and only if, h τ = 0, for τ < 0. In his quesion, h =, for < < 0, which means his sysem is no causal.

3 Fall 08 EE 33 Homework 8 soluion The Universiy of Texas a Ausin [ ] y () = x( τ) h ( τ) dτ = u( τ + ) u ( τ + ) u ( τ ) dτ In order o calculae he convoluion we should break ime domain ino hree regions. s case (No overlap): for + < < 3 In his case, x τ and h( τ) do no face any overlap, so y = 0 [ ] u( τ + ) u( τ + ) u( τ ) dτ = nd case (parial overlap): for 3 < here is parial overlap < < beween x τ and h( τ) + + [ ] y () = u( τ + ) u ( τ + ) u ( τ ) dτ = dτ = τ = ( + ) ( ) = rd case (complee overlap): for + + y ( ) = dτ = τ = ( + ) ( ) = 4 + Therefore: 0, < 3 y () = + 3, 3 < 4, MATLAB code for ploing oupu: fs = 8000; = -5: /fs :4; yy = zeros(size()); yy (>=-3 & <) = (>=-3 & <)+3; %second case -3 =< < and y() = +3 yy( >= ) = 4; % hird case >= and y() = 4 plo(,yy) ylim ([ ]) xlabel ('(s)') ylabel ('y()')

4 Fall 08 EE 33 Homework 8 soluion The Universiy of Texas a Ausin Problem 3: x () = u () 3 3 () ()* () ( ) ( ) 3 τ τ = = τ δ τ ( τ) τ = ( τ) δ( τ) τ 3 ( τ) ( τ) τ = () + () y x h u e u d u d e u u d y y y () = u ( τδτ ) ( ) d τ= u ( 0) δτ ( ) d τ= u () δτ ( ) d τ= u () 0, < 0 () 3τ 3τ 3 ( τ) ( τ) τ 3 ( τ) τ 3τ 3τ τ y = e u u d = e u d = 3e d = e = e, 0 y x h y y u e u e u 3 3 () = ()* () = () + () = () + () = () See graphical flip-and-slide convoluion on page 6. MATLAB code: clear all fs = 8000; = -: /fs :4; unisep = zeros(size()); unisep (>= 0) = ; % define uni sep funcion x = unisep; % define inpu x() = u() impulse = dirac(); % define dirac dela funcion idx = impulse == Inf; impulse (idx) = 4; h = impulse - 3*exp(-3*).*unisep; % h() is sysem response y= exp(-3*).*unisep; % y() = sysem's oupu for x() = u() plo(,x) ylim([-0.5.5]) xlabel ('(s)') ylabel ('x()') plo(,h) xlabel ('(s)') ylabel ('h()') plo(,y) ylim([-0.5.5]) xlabel ('(s)') ylabel ('y()')

5 Fall 08 EE 33 Homework 8 soluion The Universiy of Texas a Ausin

6 Fall 08 EE 33 Homework 8 soluion The Universiy of Texas a Ausin

7 Fall 08 EE 33 Homework 8 soluion The Universiy of Texas a Ausin Problem 4: The impulse response for he firs sysem will be calculaed by placing: x () = δ () y () = h() h () = δ () δ ( ) Where he oupu of firs LTI sysem, w(), is w () = h ()* x (), and he oupu of second LTI sysem is y () = h ()* w (). Here, wo sysems are conneced in cascade: y () = h()* w () = h()* h()* x () = h ()* x () The impulse response for he cascaded sysems is: [ ] h () = h()* h() = u ()* δ() δ( ) = u ()* δ() u ()* δ( ) = u () u ( ) MATLAB code: clear all fs = 8000; = -: /fs :4; unisep0 = >= 0; unisep = >= ; h = unisep0 - unisep; % h() is sysem response plo(,h) ylim ([-0.5.5]) xlabel ('(s)') ylabel ('h()')